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Worked Examples · Example 13.2

Q.Calculate the energy equivalent of 1 g1\ \text{g} of substance.

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The energy equivalent of mass is given by Einstein's mass-energy equivalence relation E=mc2E = mc^2. For 1 g1\ \text{g} of substance, this yields 9×1013 J9 \times 10^{13}\ \text{J}.

The core idea here is Einstein's mass-energy equivalence — one of the most profound results in physics. It tells us that mass is not just a measure of inertia or gravitational pull; it is a concentrated form of energy. The relation E=mc2E = mc^2 means that even a tiny amount of mass contains an enormous amount of energy because c2c^2 is a huge number (c=3×108 m/sc = 3 \times 10^8\ \text{m/s}).

Why does this matter? In nuclear reactions (fission, fusion) or particle-antiparticle annihilation, a small fraction of mass converts into energy. The problem asks for the complete conversion of 1 g1\ \text{g} — that is, all its mass turns into energy. This is the maximum possible energy you could extract from that mass.

Let's work through the calculation step by step.

  1. Write down the formula The energy equivalent of mass mm is:

E=mc2E = mc^2

where c=3×108 m/sc = 3 \times 10^8\ \text{m/s} is the speed of light in vacuum.

  1. Convert mass to SI units The mass is given as 1 g1\ \text{g}. In the SI system, the standard unit of mass is the kilogram. So:

m=1 g=1×10−3 kgm = 1\ \text{g} = 1 \times 10^{-3}\ \text{kg}

  1. Plug in the numbers

E=(1×10−3 kg)×(3×108 m/s)2E = (1 \times 10^{-3}\ \text{kg}) \times (3 \times 10^8\ \text{m/s})^2

First square the speed of light:

c2=(3×108)2=9×1016 m2/s2c^2 = (3 \times 10^8)^2 = 9 \times 10^{16}\ \text{m}^2/\text{s}^2

Then multiply:

E=10−3×9×1016=9×1013 JE = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}\ \text{J}

  1. Interpret the result …

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